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Nataly_w [17]
3 years ago
10

Plzzzzzz plzzzzz help will give brainlist

Physics
1 answer:
ad-work [718]3 years ago
4 0
Either the first one or the second one but i believe it’s the stricter on
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JOSEPH JOGS FROM END A TO OTHER END B OF A straight 300 m road in 2 minutes 30 seconds and then turns around and jogs 100 m back
zzz [600]
(a) The average speed from A to B would be 1.76 metre per second and the average velocity from A to B would also be 1.76 metre per second 

<span>(b) The average speed from A to C would be 1.73 metre per second and the average velocity from A to C would be 0.87 metre per second</span>
3 0
3 years ago
The percent by which the fundamental frequency changed if the tension is increased by 30 percent is ? a)-20.04% b)-40.12% c)-30%
nika2105 [10]

Answer:

Percentage increase in the fundamental frequency is

d)-14.02%

Explanation:

As we know that fundamental frequency of the wave in string is given as

f_o = \frac{1}{2L}\sqrt{\frac{T}{\mu}}

now it is given that tension is increased by 30%

so here we will have

T' = T(1 + 0.30)

T' = 1.30T

now new value of fundamental frequency is given as

f_o' = \frac{1}{2L}\sqrt{\frac{1.30T}{\mu}}

now we have

f_o' = \sqrt{1.3}f_o

so here percentage change in the fundamental frequency is given as

change = \frac{f_o' - f_o}{f_o} \times 100

% change = 14.02%

3 0
3 years ago
Gold is called noble gas why
slamgirl [31]
This is what i found.
Gold<span> is a </span>noble<span> element because "in chemistry, the </span>noble<span>metals are metals that are resistant to corrosion and oxidation in moist air (unlike most base metals)."</span>
3 0
4 years ago
Read 2 more answers
When a bicycle coasts uphill, it moves slower and slower as it climbs. Why?
astra-53 [7]

Letter B because it is gaining more potential energy as it SLOWLY climbs up the hill.

the less motion the more potential energy there is

6 0
3 years ago
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A crystalline grain of nickel in a metal plate is situated so that a tensile load is oriented along the [111] crystal direction
hoa [83]

Given:

Applied stress, \sigma = 0.45 MPa

Critical Resolved Stress,  T_{cRss}= 0.242 MPa

Solution:

a). According to the question, orientation of tensile load is along [1 1 1],

\sigma = 0.45 MPa

Now, for resolved shear stress,  \t_{Rss} along [1 0 1] within (1 1 T)

let  '\theta' be the angle between [1 1 1] and [1 0 1], then by coordinate formula:

cos\theta = \frac{1\times 1 + 1\times 0 + 1\times 1}{\sqrt{1^{2}+1^{2}+1^{2}\sqrt{1^{2}+0^{2}+1^{2}}}}

cos\theta = \sqrt{\frac{2}{3}}

let  '\phi' be the angle between [1 1 1] and [1  1 1], then by coordinate formula:

cos\phi  = \frac{1\times 1 + 1\times 1 + 1\times 1}{\sqrt{1^{2}+1^{2}+1^{2}\sqrt{1^{2}+1^{2}+1^{2}}}}

cos\phi = 1

Now, for the resolved components along [1 0 1]

\t_{Rss}  = \sigma  cos\theta cos\phi

\t_{Rss} = 0.45\times 10^{6}\times \sqrt{\frac{2}{3}}\times 1 = 0.3673 MPa

b).  For required tensile stress to produce  T_{cRss}= 0.242 MPa:

\sigma _{1} = \frac{T_{cRss}}{cos\theta  cos\phi }

\sigma _{1} = \frac{0.242\times 10^{6}}{\sqrt{\frac{2}{3}}\times 1}

\sigma _{1} = 0.2964 MPa

5 0
3 years ago
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